Question

The volume percentage of $C{l}_2$ at equilibrium in the dissociation of $PC{l}_5$ under a total pressure of $1.5atm$ is $(Kp=0.202)$ ,

• A. 74.5
• B. 36.5
• C. 63.5
• D. 26.6

Answer 1

Answer: (D)

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

$1-x$ $x$ $x$

Total moles $=1-x+x+x=1+x$

${\mathrm{k}}_{\mathrm{p}}=\frac{{\mathrm{P}}_{{\mathrm{PCl}}_3}\cdot {\mathrm{PCl}}_2}{{\mathrm{P}}_{{\mathrm{PCl}}_5}}=\frac{\left(\frac{\mathrm{X}}{1+\mathrm{x}}\right)\mathrm{P}}{\left(\frac{1-\mathrm{X}}{1+\mathrm{x}}\right)\mathrm{P}}\left(\frac{\mathrm{x}}{1+\mathrm{x}}\right)\mathrm{P}$

$k_P=\frac{x^{2}P}{1-x^2}$ $1-x^2=1$

$k_P=x^{2}P$

$x=\sqrt{\frac{kp}{P}}=\sqrt{\frac{0.202}{1.5}}=\sqrt{0.134}=0.366$

Mole $\propto$ volume at constant $P\&T$

Volume percentage $=\frac{moleofC{l}_2}{totalmoles}\times 100$

$=\frac{0.366}{1.366}\times 100=26.86\%$